Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Problem 7.1.71c

Chapter 7, Problem 7.1.71c

Power lines A power line is attached at the same height to two utility poles that are separated by a distance of 100 ft; the power line follows the curve ƒ(x) = a cosh x/a. Use the following steps to find the value of a that produces a sag of 10 ft midway between the poles. Use a coordinate system that places the poles at x = ±50.
c. Use your answer in part (b) to find a, and then compute the length of the power line.

Verified step by step guidance

Recall that the power line follows the curve given by the catenary function $f(x) = a \cosh\left(\frac{x}{a}\right)$. The poles are located at $x = \pm 50$, and the sag (the vertical drop at the midpoint) is 10 ft.

From part (b), you should have an expression relating the sag to $a$, typically using the fact that the sag is the difference between the height at the poles and the height at the midpoint. This can be written as: $f(50) - f(0) = 10$, which becomes $a \cosh\left(\frac{50}{a}\right) - a = 10$.

Use the equation from step 2 to solve for $a$. This involves isolating $a$ and may require numerical methods or iterative approximation since $a$ appears both inside and outside the hyperbolic cosine function.

Once you have found the value of $a$, compute the length $L$ of the power line between the poles. The length of a curve $y = f(x)$ from $x = -50$ to $x = 50$ is given by the arc length formula: $L = \int_{-50}^{50} \sqrt{1 + \left(f'(x)\right)^2} \, dx$.

Calculate the derivative $f'(x)$ of the catenary function: $f'(x) = \sinh\left(\frac{x}{a}\right)$. Substitute $f'(x)$ into the arc length integral and evaluate the integral to find the length of the power line.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Catenary Curve and Hyperbolic Cosine Function

The power line forms a catenary curve described by ƒ(x) = a cosh(x/a), where 'a' controls the curve's shape. Understanding the hyperbolic cosine function and its properties is essential to model the sag and shape of the hanging cable accurately.

Boundary Conditions and Sag Calculation

Using the given positions of the poles at x = ±50 and the sag of 10 ft at the midpoint, you apply boundary conditions to relate 'a' to the sag. This involves setting up equations based on the curve's values at specific points to solve for the parameter 'a'.

Arc Length of a Curve

To find the length of the power line, you calculate the arc length of the catenary between the poles. This requires integrating the square root of 1 plus the derivative squared, ∫√(1 + (ƒ'(x))²) dx, over the interval from -50 to 50.

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Arc Length of Parametric Curves

Related Practice

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Acceleration, velocity, position Suppose the acceleration of an object moving along a line is given by a(t) = -k v(t), where k is a positive constant and v is the object's velocity. Assume the initial velocity and position are given by v(0) = 10 and s(0) = 0, respectively.

c. Use the fact that dv/dt = (dv/ds)(ds/dt) (by the Chain Rule) to find the velocity as a function of position.

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Definitions of hyperbolic sine and cosine Complete the following steps to prove that when the x- and y-coordinates of a point on the hyperbola x² - y² = 1 are defined as cosh t and sinh t, respectively, where t is twice the area of the shaded region in the figure, x and y can be expressed as

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b. In Chapter 8, the formula for the integral in part (a) is derived:

∫ √(z² − 1) dz = (z/2)√(z² − 1) − (1/2) ln|z + √(z² − 1)| + C.

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b. Show that ∫ sech x dx = sin⁻¹ (tanh x) + C. (Hint: Show that sech x = sech² x / √(1 − tanh² x) and then make a change of variables.)